Understanding step in derivation of convex conjugate

Question

I'm reading the section on Fenchel's Duality Theorem in Barbu and Precupanu's Convexity and Optimization in Banach Spaces, and was reading the derivation of the dual problem for a special class of problems where the perturbations are generated by translations. The primal problem is defined as:
$$ \min\{f(x)-g(Ax); x\in X\}$$
where $X$ and $Y$ are real Banach spaces, $f: X\to ]-\infty, +\infty]$ is a proper, convex, and lower-semicontinuous function, $g: Y\to [-\infty, +\infty[$ is a proper, concave, and upper-semicontinuous function and $A:X\to Y$ is a linear continuous operator. They define the perturbation function $F:X\times Y \to \overline{\mathbb{R}}$ by $F(x,y)=f(x)-g(Ax-y)$. So far, so good. The only part of the derivation I am having issues following is from $(1)\to(2)$ when they begin to determine the conjugate of $F$:
$$ \begin{aligned}
F^*(x^*,y^*) &= \sup_{(x,y)\in X\times Y}\{(x,x^*)+(y,y^*)-f(x)+g(Ax-y)\}\quad (1)\\
&= \sup_{x\in X}\sup_{z\in Y}\{ (x,x^*)+(Ax,y^*)-f(x)+g(z)-(z,y^*)\} \quad (2)\\
\end{aligned}$$
I understand that $(1)$ is just the definition of convex conjugate, and the rest of the derivation after this point: they use properties of $\sup$ and $\inf$, the adjoint of $A$, and then definition of the convex conjugate to arrive at the final form. Sorry if this isn't the best question, but going from $(1)\to(2)$, how do they "split up" $g$ as they do, and where is $z$ coming from? It makes sense in context later in the derivation when they take the supremum over $z\in Y$ to then use the definition of the convex conjugate for $g$ to arrive at the final form. I suspect that I'm missing something painfully obvious. Any help showing how to get to $(2)$ from $(1)$ would be greatly appreciated 🙂

Answer 1

It's a substitution of variables, exchanging $y$ for $z = Ax - y$. Note that, for a fixed value $x$, the map $y \mapsto Ax - y$ is surjective, and hence $$\sup_{z \in Y} \{(Ax - z, y^*) + g(z)\} = \sup_{y \in Y} \{ (y, y^*) + g(Ax - y)\}.$$ Then, \begin{aligned} F^*(x^*,y^*) &= \sup_{(x,y)\in X\times Y}\{(x,x^*)+(y,y^*)-f(x)+g(Ax-y)\} \\ &= \sup_{x\in X}\sup_{y\in Y}\{(x,x^*)+(y,y^*)-f(x)+g(Ax-y)\} \\ &= \sup_{x\in X}\sup_{z\in Y}\{ (x,x^*)+(Ax - z,y^*)-f(x)+g(z)\} \\ &= \sup_{x\in X}\sup_{z\in Y}\{ (x,x^*)+(Ax,y^*)-f(x)+g(z)-(z,y^*)\}. \end{aligned}